Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

  1. $p^2 - 4pq$
  2. $2pq$
  3. $p^2 - pq$
  4. $4pq$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf one root of the equation $ax^{2}+bx+c=0$ is the square of the other, then

  1. $b^{2}+ac^{2}+a^{2}c=3abc$
  2. $b^{3}+ac^{2}+a^{2}c=3abc$
  3. $b^{2}+ac^{2}+a^{2}c+3abc=0$
  4. $b^{3}+ac^{2}+a^{2}c+3abc=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $z _{1}$ is a root of the equation $a^{n} _{0}z^{n}+a _{1}z^{n-1}+....+a _{n-1^{z}}+a _{n}=3$, where $|a _{i}|<2$ for $i=0,1,....,n.$ Then,

  1. $|z _{1}|>\dfrac {1}{3}$
  2. $|z _{1}|<\dfrac {1}{4}$
  3. $|z _{1}|>\dfrac {1}{4}$
  4. $|z|<\dfrac {1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to question,

${l} { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } }=3 \ \Rightarrow \left| { { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } } } \right| =\left| 3 \right|  \ \Rightarrow \left| { { a _{ 0 } } } \right| \, { \left| z \right| ^{ n } }+\left| { { a _{ 1 } } } \right| \, { \left| z \right| ^{ n-1 } }\, +..........+\left| { { a _{ n-1 } } } \right| \, \left| z \right| \, +\left| { { a _{ n } } } \right| \ge 3 \ \Rightarrow 2\, ({ \left| z \right| ^{ n } }+{ \left| z \right| ^{ n-1 } }+...........\left| z \right| +1)\, \, >\, 3 \ \Rightarrow (1+\left| z \right| +{ \left| z \right| ^{ 2 } }+...........+{ \left| z \right| ^{ n } })\, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow \frac { { \, \, \, \, 1-{ { \left| z \right|  }^{ n+1 } } } }{ { 1-\left| z \right|  } } \, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow 2-2{ \left| z \right| ^{ n+1 } }\, >\, 3-3\left| z \right|  \ \Rightarrow 2{ \left| z \right| ^{ n+1 } }<3\left| z \right| \, -1 \ \Rightarrow 3\, \left| z \right| -1\, >0 \ \, \, \, \, \, \therefore \, \, \, \left| z \right| \, >\, \frac { 1 }{ 3 }  \ so\, the\, correct\, option\, is\, \, A$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $p, q, r, s, t$ are the roots of the equation $x^5-1 = 0$, then $p^{ 10 }+q^{ 10 }+{ r }^{ 10 }+{ s }^{ 10 }+t^{ 10 }=$

  1. $0$
  2. $1$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p, q, r, s, t$ are all fifth root of unity
$\Rightarrow p^5=q^5= r^5= s^5= t^5=1$   
$ \Rightarrow p^{10}=q^{10}= r^{10}= s^{10}= t^{10}=1$
Hence, the required sum is $5$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The maximum number of real root of the equation $\displaystyle x^{2n} - 1 = 0$ is

  1. $\displaystyle 2$
  2. $\displaystyle 3$
  3. $\displaystyle n$
  4. $\displaystyle 2n$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2n}=1$
Now if  $n$ is odd we have
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
$x^{n}=-1$
$x=-1$
Now if $n$ is odd
$x^{n}-1=0$
$(x-1)(1+x+x^{2}+..x^{n-1})=0$
Hence $x=1$ and  the equation $1+x+x^{2}+..x^{n-1}=0$ gives $nth$  roots of unity.
Hence at most $2$ real roots.
Similarly if $n$  is even.
Then
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
Now 
$x^{n}=-1$ will given imaginary roots.
$x^{n}=1$ can be further simplified in to
$x^{\frac{n}{2}}=\pm1 $ and so on.
Hence we will get remaining pairs of imaginary roots ans two real roots $1$ and  $-1$  at the end.
Hence at-most $2$ real roots.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The roots of the equation  $z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$   are given by

  1. $-1$
  2. $\displaystyle -\frac{1}{2}+\frac{i\sqrt{3}}{2}$
  3. $\displaystyle \frac{1}{2}+\frac{i\sqrt{3}}{2}$
  4. $\displaystyle \frac{-1-i\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$\displaystyle { z }^{ 5 }+{ z }^{ 4 }+{ z }^{ 3 }+{ z }^{ 2 }+z+1=0\ \Rightarrow \frac { { z }^{ 6 }-1 }{ z-1 } =0\ \Rightarrow z\neq 1\quad &amp; \quad { z }^{ 6 }=1=\cos { 0 } +i\sin { 0 } \ \Rightarrow z={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \frac { 2k\pi  }{ 6 }  } +i\sin { \frac { 2k\pi  }{ 6 }  } \ \Rightarrow z=\cos { \frac { k\pi  }{ 3 }  } +i\sin { \frac { k\pi  }{ 3 }  } $
where $k=0,1,2,3,4,5$.

For $k=0$,
$z=1$ but $z\neq 1$

For $k=1$,
$\displaystyle z=\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } =\frac { 1+i\sqrt { 3 }  }{ 2 } $

For  $k=2$,
$\displaystyle z=\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =\frac { -1+i\sqrt { 3 }  }{ 2 } $

For  $k=3$,
$\displaystyle z=\cos { \frac { 3\pi  }{ 3 }  } +i\sin { \frac { 3\pi  }{ 3 }  } =-1$

For  $k=4$,
$\displaystyle z=\cos { \frac { 4\pi  }{ 3 }  } +i\sin { \frac { 4\pi  }{ 3 }  } =\frac { -1-i\sqrt { 3 }  }{ 2 } $

For $k=4$,
$\displaystyle z=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } =\frac { 1-i\sqrt { 3 }  }{ 2 } $

Hence, all the options A,B,C and D are correct.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\alpha,\ \beta,\ \gamma$ and $\Delta $ are the roots of the equation $x^{4}-1=0$, then the value of $\displaystyle \frac{a\alpha+b\beta+c\gamma+d\Delta}{a\gamma+b\Delta +c\alpha+d\beta}+\frac{a\gamma+b\Delta +c\alpha+d\beta}{a\alpha+b\beta+c\gamma+d\Delta }$ is

  1. $ 3\beta$
  2. $0$
  3. $ 2\gamma$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Clearly,
$\alpha = e^{i0} = 1$
$\beta = e \frac{i2\pi}{4} = i$
$\gamma = e \frac{i4\pi}{4} = -1$
$\delta = e \frac{i6\pi}{4} = -i$
So, $\dfrac {a\alpha+b\beta+c\gamma+d\delta}{ a\gamma+b\delta+c\alpha+d\beta}=\dfrac {a+bi-c-di}{- a-bi+c+di}$
$=-1$
Similarly second expression is nothing but reciprocal of first
$=\frac{1}{-1}=-1$
Ans $ = -1-1 = -2$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The number of roots of the equation $z^{15}=1$ satisfying $|\arg(z)|<\pi/2$ is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For the nth root of unity of complex number $z$ i.e. $z^n = 1$, there are 'n' total roots.
In the present case, $n=15$, thus, we have 15 roots.

$|arg(z)|<\cfrac {\pi}{2}$ 
$\Rightarrow -\cfrac {\pi}{2} < arg(z) < \cfrac {\pi}{2}$

Each root is at equal angular distance i.e. $\dfrac{2\pi}{15}$      ...(because $\dfrac{2\pi}{n}$).

$\therefore$ between $-\cfrac {\pi}{2}$ and $\cfrac {\pi}{2}$, their will be 7 roots (imcluding 1).
Hence, the correct option is B. 
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

Find all those roots of the equation $z^{12} - 56z^6 - 512 = 0$ whose imaginary part is positive.

  1. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  2. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2, $$2^{1/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  3. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{1/3} \left ( -cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( -cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( -cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  4. $2, 2 \left ( -cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( -cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(z^{6}-28)^{2}-784-512=0$
$(z^{6}-28)^{2}=1296$
$z^{6}-28=\pm36$
$z^{6}=64$ and $z^{6}=-8$
$z^{3}=\pm8$
$z=2$ and $z=-2$ ...(i)
$z^{6}=2^{3}.e^{i(2k-1)\pi}$
$z=2^{\frac{1}{2}}(e^{i\frac{(2k-1)\pi}{6}})$ where $k=1,2,3..6$.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

lf $a=\displaystyle \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}, \alpha=a+a^{2}+a^{4}$ and $\beta=a^{3}+a^{5}+a^{6}$, then $\alpha, \beta$ are the roots of the equation

  1. $x^{2}+x+1=0$
  2. $x^{2}+x+2=0$
  3. $x^{2}+2x+2=0$
  4. $x^{2}+2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a={ e }^{ i2\pi /7 }\ a^ 7=1\ \alpha +\beta $

$=a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6\ =\dfrac{a(a^ 6-1)}{(a-1)}\ =\dfrac{(a^ 7-a)}{(a-1)}\ =\dfrac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^ 2+a^ 4)(a^ 3+a^ 5+a^ 6)\ =(a^ 4+a^ 6+a^ 7+a^ 5+a^ 7+a^ 8+a^ 7+a^ 9+a^ {10})\ =(a^ 4+a^ 6+1+a^ 5+1+a+1+a^ 2+a^ 3)\ =(3+a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6)\ =(3-1)$
$=2 $
The equation can be written as
 $x^ 2-(\alpha +\beta )x +\alpha \beta  =x^ 2+x+2$
Hence, option B is correct.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths
If the expression $z^5 =32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5 - 32)=(z - 2) (z^2-pz+4)(z^2-qz+4)$, where p > q, then the value of $p^2-  2q$
  1. $8$
  2. $4$
  3. $-4$
  4. $-8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$ { z }^{ 5 }=32 $ can be factorized as
$(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)$ where $ p>q.$
 To find the value of $ { p }^{ 2 }-2q$
 Solution,
${ z }^{ 5 }=32\ { z }^{ 5 }-32=0$
$ { z }^{ 5 }-{ 2 }^{ 5 }=0$
$ z=2$ 
Will be one of the factor of given equation.
$ \quad \quad \quad \quad \quad \quad \quad { z }^{ 4 }+{ 2z }^{ 3 }+{ 4z }^{ 2 }+8z+16\ \therefore \quad z-2\sqrt { { z }^{ 5 }-32\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  } \ \quad \quad \quad \cfrac { \pm { z }^{ 5 }\mp { 2z }^{ 4 } }{ { \quad \quad \quad \quad \quad \quad 2z }^{ 4 }-32 } \ \quad \quad \quad \quad \quad \quad \cfrac { \pm { 2z }^{ 4 }\mp 4{ z }^{ 3 } }{ \quad \quad \quad \quad \quad \quad \quad 4{ z }^{ 3 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 4{ z }^{ 3 }\mp 8{ z }^{ 2 } }{ \quad \quad \quad \quad \quad \quad \quad 8{ z }^{ 2 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 8{ z }^{ 2 }\mp 16z }{ \quad \quad \quad \quad \quad \quad \quad 16z-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 16z\mp 32 }{ 0 } \ (z-2)({ z }^{ 4 }+2{ z }^{ 3 }+4{ z }^{ 2 }+8z+16)\longrightarrow (1)\ { z }^{ 5 }-32=(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)\quad \ { z }^{ 5 }-32=(z-2)({ z }^{ 4 }+(p+q)z^{ 3 }+(8+pq){ z }^{ 2 }+16-(4p+4q)z)\longrightarrow (2)$
On comparing $ (1)&amp; (2)$ we get
$ p=-2-q\ 8+(-2-q)q=4\ -2q-{ q }^{ 2 }=-4\ { q }^{ 2 }+2q-4=0\ d=4+16=20\ q=\cfrac { -2\pm 2\sqrt { 5 }  }{ 5 } \ q=-1\pm \sqrt { 5 } \ { \parallel  }^{ rly }p=-1\pm \sqrt { 5 } $
But$\quad p>q\therefore p=-1+\sqrt { 5 } &amp; q=-1-\sqrt { 5 } $
value of ${ p }^{ 2 }-2q={ \left( -1+\sqrt { 5 }  \right)  }^{ 2 }+2{ \left( -1-\sqrt { 5 }  \right)  }\ =1+5-2\sqrt { 5 } +2+2\sqrt { 5 } \ { p }^{ 2 }-2q=8\ $

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The circle ${ x }^{ 2 }+{ y }^{ 2 }=4$ cuts the line joining the points $A(1,0)$ and $B(3,4)$ in two points P and Q. Let $\dfrac { BP }{ PA } =\alpha$ and $\dfrac { BQ }{ QA } =\beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

  1. $3{ x }^{ 2 }+2x-21=0$
  2. $3{ x }^{ 2 }+2x+21=0$
  3. $2{ x }^{ 2 }+3x-21=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The equation of line joining $A(1,0)$ and $B(3,4)$ is $\dfrac{y-0}{x-1}=\dfrac{4-0}{3-1}\implies y=2x-2$    ... (1)

The point of intersection of this line and circle are

$x^2+(2x-2)^2=4\implies x^2+4x^2+4-8x=4\implies x=0,\dfrac{8}{5}$

Hence, points of intersection are $P(0,-2)$ and $Q\left(\dfrac{8}{5},\dfrac{6}{5}\right)$

Now, $BP=\sqrt{3^2+6^2}=\sqrt{45}$, $PA=\sqrt{1^2+2^2}=\sqrt{5}$, 

$BQ=\sqrt{\left(3-\dfrac{8}{5}\right)^2+\left(4-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{245}{25}}$ and $QA=\sqrt{\left(1-\dfrac{8}{5}\right)^2+\left(0-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{45}{25}}$

$\therefore \dfrac{BP}{PA}=3=\alpha$ and $\dfrac{BQ}{QA}=\dfrac{7}{3}=\beta$

The equation with roots $\alpha$ and $\beta$ is $(x-\alpha)(x-\beta)=0\implies (x-3)(3x-7)=0\implies 3x^2-16x+21=0$

This is the required answer.
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

lf the expression $3x^{2}+2pxy+2y^{2}+2ax-4y+1$ can be resolved into two linear factors, then $p$ must be a root of the equation

  1. $x^{2}+ax+6=0$
  2. $x^{2}+4ax+6=0$
  3. $x^{2}+4ax+2a^{2}+6=0$
  4. $x^{2}-4ax+6=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $3x^{2}+2pxy+2y^{2}+2ax-4y+1=0.$, then.
$\Delta =abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
$=3(2)(1)+2(-2)(a)(p)-3(-2)^{2}-2(a)^{2}-1(p)^{2}=0$
$6-4ap-12-2a^{2}-p^{2}=0$
$p^{2}+4ap+2a^{2}+6=0$
$\Rightarrow p $ is a solution of $x^{2}+4ax+2a^{2}+6=0$